I have been working on the double integral proof for the volume of a cone. I found that I can use a unit-sphere Where the base of the cone is the equator and the height is the distance ${\rho}$ to the pole. Using area of a ${\Delta}$=${1/2bh}$ I define the Area of the ${\Delta}$ with base ${\rho}$ and height ${\rho}$ as ${1/2\rho^2}$.

It follows $$V_{cone}={\int_0^R \int_0^{2\pi}1/2\rho^2d\rho d\theta}$$ $$V_{cone}={{1/6}R^3 \int_0^{2\pi}d\theta}$$ $$V_{cone}={{1/3}\pi R^3={1/3}\pi R^2 R}$$ Since the H=R I get $${1/3\pi R^2H}$$

I'd like to be able to prove this without the unit-sphere but I am having trouble understanding the logic in the triple integral proof in cylindrical coordinates. Can anyone help me see the parallels to my approach So I can finally understand the cylindrical method? Thanks in advance.


(Moved self answer into OP to condense the thread.)

I now understand.

Start with the observation that axes are arbitrary. Next we need to find the equation of the line in a cylinder with the height on the Z axis. $${Z=mX +H}$$ Where H is the Z intercept (the apex of the cone)

Change the X axis to ${\rho}$ then the slope of the line is $${(H/R)}$$ Where H is the apex of the cone and R is the radius of the base. (the limit of the base on ${\rho}$)

THen the equation of the slant is $${Z=(H/R)\rho + H}$$ IF we assume the slope is negative then the equation becomes $${Z=H-(H\rho/R)}$$

Now we can integrate $${\int_0^R\int_0^{2\pi}\int^H_{(H/R)p}(\rho )d\rho d\theta dz}$$ Integrating over z will give you the equation of the line but the ${\rho}$ keeps you from dragging it out of the integral with respect to ${\rho}$ So now we have $${\int_0^R \int_0^{2\pi} \rho[H-(H\rho/R)]d\rho d\theta}$$ ... $${\int_0^R\int_0^{2\pi}H\rho-(H\rho^2)/R}$$.. $${\int_0^{2\pi} (HR^2)/2-(HR^2 )/3}$$... $${\int_0^{2\pi}[(3HR^2)/6-(2HR^2)/6 = \int_0^{2\pi} HR^2/6}$$... finally $${(2\pi HR^2)/6 = (R^2\pi H)/3}$$

Thanks to @AlexJordan for the link.


You don't need calculus for the volume of something like this if you take Cavalieri's principle and dimensional rescaling as givens. (If you do not, see below for my comments on that.)

Start be decomposing a $1\times1\times1$ cube into six equal square-based pyramids, using a vertex at the center of the cube.

So the volume of these pyramids are each $\frac{1}{6}$. The height of each pyramid is $\frac{1}{2}$. Now, scaling a solid by $h$ in any one dimension scales the volume by $h$. (Again, see below if you feel calculus should justify this.) So to wind up with a pyramid of height $h$, scale in the vertical direction by $2h$, giving a volume of $\frac13h$.

Now Cavalieri's principle allows you change a horizontal cross-section to any shape of the same area and the volume of the shape will not change. If its a cone that you are looking for, you can change all the cross-sectional squares to circles of area $1$ and the volume is still $\frac13h$.

Lastly you can change the dimensions on the cross-sectional shape (in this case a circle) to whatever you want, as long as you scale all of the cross sections uniformly. If the area of the base cross section becomes $B$ (say after scaling or contracting the radii by $\sqrt{B}$) then you have scaled the volume by $B$, and have $V=\frac13Bh$.

Some say that this kind of argument really relies on calculus. But ancient civilizations had these principles long before calculus was formally developed. And to say that this kind of argument is not as valid, to me, is a bit like saying you need field axioms to prove $1+1=2$.

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  • $\begingroup$ thanks for the input ( and don't get me wrong it was enlightening) but call me stubborn I still want to understand the trip int approach. $\endgroup$ – Chris Dec 20 '13 at 4:32
  • $\begingroup$ Take a look at this. $\endgroup$ – alex.jordan Dec 20 '13 at 4:51
  • $\begingroup$ I thought about that approach, but polar, cylindrical, and spherical coordinate lend themselves to much cleaner integrals. Once I understand the equation of the line in cylindrical coordinates it should just be a simple matter of $${\int_0^R \int_0^{2\pi}\int_0^H f(\rho)d\rho d\theta dz}$$ $\endgroup$ – Chris Dec 20 '13 at 4:56
  • $\begingroup$ Thanks for the link looks like what I am after. Will study tomorrow though...bed time. $\endgroup$ – Chris Dec 20 '13 at 5:00
  • $\begingroup$ Thanks again for the link it really helped me understand the full process. I was able to explain it to someone else in detail. $\endgroup$ – Chris Dec 20 '13 at 17:16

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